3

The sequence $a_n$ satisfies $$a_1=1,\\a_n+a_{n+12}=2a_{n+6},\\a_n+a_{n+14}=2a_{n+7}.$$

I know that these conditions imply that $a_n$ consists of several arithmetic subsequences with index interval as $6$ and $7.$ Because $6$ and $7$ are coprime, I guess this implies that the whole sequence is arithmetic.

But how do we prove that the sequence is indeed arithmetic?

Also: is the sequence arithmetic regardless of the value of $a_1\,?$

ryang
  • 38,879
  • 14
  • 81
  • 179
Hans J
  • 175
  • 7
  • Maybe $a_1$ is also unnecessary. I just wonder if it is arithmetic, no matter what the common difference is. – Hans J Jul 12 '22 at 02:40
  • Subtracting the two equations and reindexing gives $a_{n+1}-a_n = (a_{n+8}-a_{n+6})/2$. Might be helpful. – eyeballfrog Jul 12 '22 at 02:47
  • 1
    @HansJ Consider you can prove that each increase of the index by $6$ is equal, plus likewise for an increase by $7$, and that $a_{n+42}$ is in common between those $2$ sets of increases. – John Omielan Jul 12 '22 at 02:53
  • 1
    @JohnOmielan Can you write the process? I find it hard to prove all the subsequences have the same common difference. – Hans J Jul 12 '22 at 16:56
  • @HansJ As you requested, I've written a quite detailed answer. I've used a lot of variables to not assume anything but, instead, prove they are all equal to each other. Also, I hope you find my note at the end useful to understand how this can be generalized, and that your concern about how the recurrence relations interact is valid since we need the two index differences, e.g., $6$ and $7$, to be coprime for the entire sequence $a_n$ to always be an arithmetic sequence. – John Omielan Jul 12 '22 at 18:32

3 Answers3

2

The $a_n$ sequence is always arithmetic with the two provided recurrence relations of

$$a_{n} + a_{n+12} = 2a_{n+6} \tag{1}\label{eq1A}$$

$$a_{n} + a_{n+14} = 2a_{n+7} \tag{2}\label{eq2A}$$

Choose any $n = n_1 \ge 1$. There exist $e_1$ and $e_2$ with $a_{n_{1}+6} = a_{n_{1}} + 6e_{1}$ and $a_{n_{1}+12} = a_{n_{1}} + 12e_{2}$. Using these in \eqref{eq1A} gives

$$\begin{equation}\begin{aligned} a_{n_{1}} + (a_{n_{1}} + 12e_{2}) & = 2(a_{n_{1}} + 6e_{1}) \\ 2a_{n_{1}} + 12e_{2} & = 2a_{n_{1}} + 12e_{1} \\ e_{2} & = e_{1} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

This means $a_{n_{1} + 12} = a_{n_{1}} + 12e_{1} \; \to \; a_{n_{1} + 12} - a_{n_{1} + 6} = a_{n_{1} + 6} - a_{n_{1}} = 6e_{1}$, i.e., the change in values of $a_n$ when the index increases by $6$ is the same. Note we can repeat this using $n = n_1 + 6$ to get that $a_{n_{1} + 18} - a_{n_{1} + 12} = 6e_{1}$, as well as going up or down by $6$ any number of times, as can be proven using induction. This means for all integers $k_1$ with $n_1 + 6k_1 \ge 1$, we've that

$$a_{n_1 + 6k_1} = a_{n_1} + 6k_{1}e_{1} \tag{4}\label{eq4A}$$

We can use a similar procedure with \eqref{eq2A} to get that, for some $f_1$, we have for all integers $k_2$ with $n_1 + 7k_2 \ge 1$ that

$$\color{blue}{a_{n_1 + 7k_2} = a_{n_1} + 7k_{2}f_{1}} \tag{5}\label{eq5A}$$

Using $k_1 = 7$ in \eqref{eq4A} and $k_2 = 6$ in \eqref{eq5A} gives

$$a_{n_1 + 42} = a_{n_1} + 6(7)e_1 = \color{blue}{a_{n_1} + 7(6)f_1} \; \; \to \; \; e_1 = f_1 = d_1 \tag{6}\label{eq6A}$$

We can next use $n = n_1 + 7$ as a starting point, and repeat the procedure above to get that there exists $e_{3}$ and $f_{3}$ where, for any integers $k_3$ and $k_4$, we have that

$$\color{red}{a_{n_1 + 7 + 6k_3} = a_{n_1 + 7} + 6k_{3}e_{3}} \tag{7}\label{eq7A}$$

$$a_{n_1 + 7 + 7k_4} = a_{n_1 + 7} + 7k_{4}f_{3} \tag{8}\label{eq8A}$$

with $e_{3} = f_{3}$. Since \eqref{eq8A} has a base point of $n_1 + 7$, and \eqref{eq5A} applies for all $k_2$, we get $f_{3} = f_{1}$. An alternate way to show this is to use $k_2 = k_4 + 1$ and $k_2 = 1$ in \eqref{eq5A} to therefore get from \eqref{eq8A} that

$$\begin{equation}\begin{aligned} a_{n_1 + 7(k_4 + 1)} & = a_{n_1 + 7} + 7k_{4}f_{3} \\ \color{blue}{a_{n_1} + 7(k_4 + 1)f_{1}} & = (\color{blue}{a_{n_1} + 7f_{1}}) + 7k_{4}f_{3} \\ 7k_{4}f_{1} & = 7k_{4}f_{3} \\ f_{1} & = f_{3} \end{aligned}\end{equation}\tag{9}\label{eq9A}$$

i.e., $e_{3} = f_{3} = d_{1}$ as well. Thus, using $k_3 = -1$ in \eqref{eq7A}, plus $k_2 = 1$ in \eqref{eq5A}, we next have

$$\begin{equation}\begin{aligned} \color{red}{a_{n_{1} + 7 + 6(-1)}}\text{ } & \color{red}{= a_{n_1 + 7} + 6(-1)d_{1}} \\ a_{n_{1} + 1} & = (\color{blue}{a_{n_1} + 7(1)d_{1}}) - 6d_{1} \\ a_{n_{1} + 1} & = a_{n_1} + d_{1} \end{aligned}\end{equation}\tag{10}\label{eq10A}$$

Note we can use a procedure similar to the above with a starting point of $n = n_1 + 7k$ for any integer $k$ where $n \ge 1$ (actually, it's also true for $n \le 0$ as well) to get that

$$a_{n_{1} + k} = a_{n_1} + kd_{1} \tag{11}\label{eq11A}$$

i.e., $a_{n}$ is an arithmetic sequence with a common difference of $d_1$. This shows the value of $a_{1}$ doesn't affect whether or not it's an arithmetic sequence but, instead, just determines the actual values of $a_n$ based on the multiples of $d_1$ offsets from $a_1$.

Note: This works not only for index differences of $6$ and $7$, but also for any $2$ integers which are coprime to each other. The basic underlying reason for this is Bézout's identity which states that, if the index differences were $x$ and $y$ with $\gcd(x,y) = d$, then there are integers $m$ and $n$ with

$$mx + ny = d \tag{12}\label{eq12A}$$

This allows matching up the value differences with index offsets of $d$ for the sequences determined by integer index differences of $x$ and $y$ to each other, as done above for $x = 6$ and $y = 7$ with $d = 1$. However, if $d \gt 1$, then the best we can state is that there are multiple (possibly different) arithmetic sequences where, for each $0 \le i \le d - 1$, we have $a_{kd + i}$ forms an arithmetic sequence for all integers $k$.

John Omielan
  • 47,976
  • 1
    Thanks, it's great! In my understanding, firstly, you choose $n_1$ as the base to get Eq. (4) and (5), then choose $n_1+7$ as the base to get Eq. (7) and (8). By doing this, what you prove is, the $n_1$-based 6-subsequence has the same common difference as $(n_1+7)$-based 6-subsequence (i.e. $(n_1+1)$-based 6-subsequence). Therefore, by induction, we know for any starting point $n_1$, the 6-subsequence has the same common difference. Now what remains to prove is two subsequence leading by different starting point (mod 6) can be merged into one arithmetic sequence, which is what Eq. (10) does. – Hans J Jul 17 '22 at 03:13
  • @HansJ You're welcome. I believe your understanding is correct. The main issue is to show the $6$-subsequence and $7$-subsequences differences, per each index change, are the same, which I did via those equations as you describe. Also, a key for your particular case is, as you state, that $n_1 + 7 \equiv n_1 + 1 \pmod{6}$. Nonetheless, as I explain in my later note, the main key for the general case is that the $2$ subsequence lengths are coprime, with it being somewhat more complicated (but not too much) if these values are not just $1$ different from each other, like $6$ and $7$ are. – John Omielan Jul 17 '22 at 03:21
  • The key is $n_1+7\equiv n_1+1(\mathbf{mod} \quad 6)$, which can cover all $n\in \mathbb{N}^$ from $n_1$ by induction. If the two relations are not 6 and 7, however, it is , e.g. 6 and 8 instead, then because $n_1+8\equiv n_1+2 (\mathbf{mod}\quad 6)$, we can not get all $n\in \mathbb{N}^$ from any $n_1$ by induction. – Hans J Jul 17 '22 at 03:26
  • @HansJ That is correct. With your example of $8$ and $6$, since $\gcd(8,6) = 2$, the best that can be proven is that all of the odd index values form an arithmetic sequence among themselves, and likewise all of the even index values together form an arithmetic sequence, with them possibly being different (e.g., having the same common difference but being offset from each other or, alternatively, having different common differences) although, obviously, they could also form together one entire arithmetic sequence. I believe you have a good understanding of how this works. – John Omielan Jul 17 '22 at 03:31
  • 1
    Thanks a lot, I think for any $(p,q)$, what we should prove is, $\forall k_1\in \mathbb{N}$, $\exists k_2\in \mathbb{N}$, s.t. $n_1+q k_2\equiv n_1+k_1 (\mathbf{mod} \quad p)$, i.e. $\forall k_1\in \mathbb{N}$, $\exists k_2\in \mathbb{N}$, s.t. $q k_2\equiv k_1 (\mathbf{mod} \quad p)$ – Hans J Jul 17 '22 at 04:16
  • 1
    @HansJ You're welcome. If $\gcd(p,q) = d =1$, then $q$ has an inverse modulo $p$, so your statement becomes $qk_2 \equiv k_1 \pmod{p} \iff k_2 \equiv q^{-1}k_1 \pmod{p}$. If $d \gt 1$, then with $k_1 = 1$, we get $qk_2 \equiv 1 \pmod{p} ; \to ; p \mid qk_2 - 1$, but $d \mid p$ and $d \mid q$ means $d \mid 1$, which is not possible. Thus, with your earlier example of $6$ and $8$, only values with the same parity as $n_1$ may be reached, i.e., if $n_1$ is odd, then only odd index values may be reached and, if $n_1$ is even, then only even indices. Finally, note \pmod{p} gives "$\pmod{p}$". – John Omielan Jul 17 '22 at 20:32
  • @HansJ Yes, my $q^{-1}$ is meant to be the inverse modulo $p$, with it being the same as your $q'$. Note it's fairly common, and at least somewhat standard, when dealing with congruences to have $q^{-1}$ not mean the rational fraction $\frac{1}{q}$ but, rather, an integer (fairly often, at least implicitly, with it being $0 \lt q^{-1} \lt p$) where $q(q^{-1}) \equiv 1 \pmod{p}$. Apart from you writing $q'k_2$ instead of $q'k_1$ in a couple of places, everything else in your comment is correct. – John Omielan Jul 18 '22 at 05:28
  • Wonderful! Now I get the clue. But is there a problem in "$q$ has an inverse modulo $p$"? I think the inverse modulo of $q$ is the $q^{-1}$ in your comment. If $\gcd(p,q)=d=1$, then there exists $q'\in \mathbb{N}$, s.t. $qq'\equiv 1\pmod{p}$. Then $qk_2\equiv k_1 \pmod{p}\Leftrightarrow qq'k_2\equiv q'k_1 \pmod{p}\Leftrightarrow k_2\equiv q' k_1\pmod{p}$. Then for any $k_1$, choose $k_2=q'k_1+kp$. – Hans J Jul 18 '22 at 05:52
  • @HansJ My response to your comment is just before it, but with you now having fixed those couple of typos. I believe that, practically (especially for particularly large $p$ and $q$), it's generally better with $p$ and $q$ formula index change values to use my $(12)$ to get $m$ and $n$ so $mp + nq = 1$, such as by using the extended Euclidean algorithm or, for simple enough cases, just inspection. ... – John Omielan Jul 19 '22 at 01:30
  • @HansJ (cont.) An example where the latter method can be used is your $p = 7$ and $q = 6$, where we have just simply $1(7) + (-1)(6) = 1$. Another one is with $p = 5$ and $q = 3$, where $2(5) + (-3)(3) = 1$. Basically, we can use one of those values ( (e.g., $p$) multiple times to increase the index and then other value (e.g., $q$) additional times to decrease the index to end up just being $1$ ahead. Then for any $k_1$, we can multiply both sides by $k_1$ to get $(mk_1)p + (nk_1)q = k_1$. – John Omielan Jul 19 '22 at 01:31
0

Please ignore this flawed proof -- see the comments by @Hans J.

$$a_n + a_{n+12} = 2a_{n+6} \Longleftrightarrow$$ $$ a_{n+12}-a_{n+6} = a_{n+6} -a_n \Longleftrightarrow$$ $$ \exists p: a_{6m+i} = a_i + pm\Longleftrightarrow$$ $$ a_{i-6m} = a_i - pm$$ Similarly: $$ \exists q: a_{7m+j} = a_j + qm$$

$$a_{m+1} = a_{7m-6m+1} = a_{7m+1} - pm = a_1 + qm - pm \Longrightarrow$$ $$a_{m+1} = a_1 + (q-p)m$$

I.e., sequence $\{a_i\}$ is an arithmetic progression with a common difference of $q-p$.

blamocur
  • 1,153
  • 2
    There are some problems. In $\exists p: a_{6m+i}=a_i+pm$, the $p$ depends on the $i$. The given condition cut the whole ${a_i}$ into 6 arithmetic subsequences with initial term $a_1,a_2,\cdots, a_6$. The subsequences with different initial terms may have different common differences. So in $a_{m+1}=a_{7m-6m+1} = a_{7m+1}-pm$, the $p$ depends on $7m+1$ according to the residue of $(7m+1)/6$. So maybe with different $m$, we have different $p$ then different $p-q$. – Hans J Jul 12 '22 at 16:54
  • Good catch! Back to the drawing board :( – blamocur Jul 12 '22 at 18:02
0

(Too long for comment)

Eliminating $a_n$ gives the relation

$$a_{n+14} - a_{n+12} = (a_{n+14} - a_{n+13}) + (a_{n+13} - a_{n+12}) = 2(a_{n+7} - a_{n+6})$$

If $b_n=a_{n+1}-a_n$ is the sequence of first-order differences of $a_n$, then this is equivalent for $n\ge1$ to

$$b_{n+13}+b_{n+12}=2b_{n+6} \implies b_{n+7} + b_{n+6} - 2b_n = 0$$

with solution

$$b_n = c_1 + \sum_{i=2}^7 c_i {r_i}^n$$

where $r_i$ are the roots to the characteristic polynomial

$$r^7 + r^6 - 2 = (r-1) (r^6 + 2r^5 + 2r^4 + 2r^3 + 2r^2 + 2r + 2)$$

Solving for $a_n$ by substitution gives

$$\begin{align*} a_n &= a_1 + \sum_{i=1}^n b_i \\[1ex] &= 1 + \sum_{i=1}^n \left(c_1 + \sum_{j=2}^7 c_j {r_j}^n\right) \\[1ex] &= 1 + c_1n + \left(c_2 \rho^n + c_3 \bar\rho^n + c_4 \sigma^n + c_5 \bar\sigma^n + c_6 \tau^n + c_7 \bar\tau^n\right)n \end{align*}$$

where $\rho,\sigma,\tau$ and their conjugates $\bar\rho,\bar\sigma,\bar\tau$ are the remaining complex characteristic roots.

Then it suffices to show the exponential terms will cancel or reduce to a constant to be combined with $c_1$. This would be trivial if we knew $\gcd(c_{2i},c_{2i+1})\neq1$ for $i\in\{1,2,3\}$...

user170231
  • 19,334